Question

If $f\left(x\right) = \frac{\sin\left[x\right]}{\left[x\right]} , \left[x\right]\ne0 $
= 0 , [ x ] = 0
Where [x] denotes the greatest integer less than or equal to $x$. then $\displaystyle \lim_{x \to 0} f(x) $ equals -

• A. 1
• B. 0
• C. -1
• D. none of these

Answer 1

Answer: (D)

The given function can be restated as $ f(x) = \begin{cases} \frac{\sin [x]}{[x]} , & \quad \text{if } x \in (- \infty , 0) \cup [1, \infty ] \\ 0 & \quad \text{if } x \in [0 , 1] \end{cases}$
$ \therefore \displaystyle \lim_{x \to 0^{-} } f\left(x\right) =\displaystyle \lim_{h \to 0} \frac{\sin\left[-h\right]}{\left[-h\right]}$
$ =\displaystyle \lim_{x \to 0^{-}} \frac{\sin\left(-1\right)}{\left(-1\right)} = \sin 1$
And $\displaystyle \lim_{x\to 0^{+}} f\left(x\right) =\displaystyle \lim_{h \to 0} 0 = 0$
$ \because \displaystyle \lim_{x \to 0^{-} } f\left(x\right) \ne \displaystyle \lim_{x \to 0^{+} } f\left(x\right) $ as $\sin 1 \ne 0$
$ \therefore \displaystyle \lim_{x \to 0 } f\left(x\right) $ does not exist.